HDU 4908 BestCoder Sequence——BestCoder Round #3

BestCoder Sequence

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

Mr Potato is a coder.
Mr Potato is the BestCoder.

One night, an amazing sequence appeared in his dream. Length of this sequence is odd, the median number is M, and he named this sequence as Bestcoder Sequence.

As the best coder, Mr potato has strong curiosity, he wonder the number of consecutive sub-sequences which are bestcoder sequences in a given permutation of 1 ~ N.

 

Input

Input contains multiple test cases. 
For each test case, there is a pair of integers N and M in the first line, and an permutation of 1 ~ N in the second line.

[Technical Specification]
1. 1 <= N <= 40000
2. 1 <= M <= N

 

Output

For each case, you should output the number of consecutive sub-sequences which are the Bestcoder Sequences. 

 

Sample Input

1 115 34 5 3 2 1

 

Sample Output

13
Hint
For the second case, {3},{5,3,2},{4,5,3,2,1} are Bestcoder Sequence.
 

 

Source

BestCoder Round #3

 

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出题人的解题思路：

将大于M的数标记为1，小于M的数标记为-1，M本身标记为0，则题目就是要求和为0并且包括M的连续序列的个数；用sum_i表示从第一个数到第i个数的标记的和，对于所有大于等于M的位置的i，我们要求小于M的位置的sum_j == sum_i的个数的和即为答案。

该题简单来说其实就是给你一个序列，里面是1~N的排列，问你其中以M为中位数的奇数长度的区间有多少个

刚开始没有理解median number的意思，以为M要在区间的正中间，看了Clarification之后才发现是中位数的意思。

而M要是一个区间内的中位数，需要满足的是该区间内比M大的数的个数与比M小的数的个数相等

那么我们可以记录M左右两边区间比M大的和比M小的差值的个数，再相乘。
要想做到这一点，可以先遍历M的左边，记录差值个数，而遍历右边时直接加上与左边差值一样的即可。

比如说记录包含M的区间中M左侧有多少个比M大的数或比M小的数，比M大的和比M小的可以抵消，记录净差值

若仍不明白，可以提出，我会给以回复，并对解题报告作出修正，谢谢

放上一组测试数据，以供参考

Input

7 4
1 5 4 2 6 3 7
Output
8

Hint
{4}
{1,5,4}
{5,4,2}
{4,2,6}
{1,5,4,2,6}
{4,2,6,3,7}
{5,4,2,6,3}
{1,5,4,2,6,3,7}

#pragma comment(linker, "/STACK:1024000000,1024000000")#include<stdio.h>#include<string.h>#include<stdlib.h>#include<queue>#include<math.h>#include<vector>#include<map>#include<set>#include<stdlib.h>#include<cmath>#include<string>#include<algorithm>#include<iostream>#define exp 1e-10#define ll __int64using namespace std;const int N = 20005;const int inf = 1000000000;const int mod = 258280327;int a[2*N],s[2*N];int main(){    int n,m,i,k,j,ans;    while(~scanf("%d%d",&n,&m))    {        memset(s,0,sizeof(s));        for(i=0;i<n;i++)        {            scanf("%d",&a[i]);            if(a[i]==m)                k=i;        }        for(i=k,j=0;i>=0;i--)        {            if(a[i]!=m)                a[i]>m?j++:j--;            s[N-j]++;        }        for(i=k,ans=j=0;i<n;i++)        {            if(a[i]!=m)                a[i]>m?j++:j--;            ans+=s[N+j];        }        printf("%d\n",ans);    }    return 0;}

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